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kap26 [50]
3 years ago
7

What polyhedron can be formed from a net diagram that had 2 congruent hexagons and 6 congruent rectangles

Mathematics
2 answers:
Sidana [21]3 years ago
4 0
Hello,
an hexagonal  right prisme.

DiKsa [7]3 years ago
4 0

Answer:

Hexagonal prism

Step-by-step explanation:

Polyhedron is a three-dimensional figure.

The name of polyhedron with respect to the number sides.

Let's see the name of the polyhedron with respect to the number sides.

cube - which is formed by 6 square faces

Pentagonal prism - which is formed by 2 congruent pentagons and 5 congruent rectangles

Hexagonal prism - which is formed by 2 congruent hexagons and 6 congruent rectangles.

Hexagon is a sided figure.

Therefore, the answer is "Hexagonal prism".

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Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tab
Leona [35]

Answer:

a. 5 b. y = -\frac{3}{4}x + \frac{1}{2} c. 148.5 d. 1/7

Step-by-step explanation:

Here is the complete question

Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(πx) + x² +3. (a) Find f" (-2) (b) Write an equation for the line tangent to the graph of y = 1/f(x) at x = 0. (c) Let I be the function defined by g(x) = f (√(3x² + 4). Find g(2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.

Solution

a. f"(2)

f"(x) = df'(x)/dx = d(sin(πx) + x² +3)/dx = cos(πx) + 2x

f"(2) = cos(π × 2) + 2 × 2

f"(2) = cos(2π) + 4

f"(2) = 1 + 4

f"(2) = 5

b. Equation for the line tangent to the graph of y = 1/f(x) at x = 0

We first find f(x) by integrating f'(x)

f(x) = ∫f'(x)dx = ∫(sin(πx) + x² +3)dx = -cos(πx)/π + x³/3 +3x + C

f(0) = 2 so,

2 = -cos(π × 0)/π + 0³/3 +3 × 0 + C

2 = -cos(0)/π + 0 + 0 + C

2 = -1/π + C

C = 2 + 1/π

f(x) = -cos(πx)/π + x³/3 +3x + 2 + 1/π

f(x) = [1-cos(πx)]/π + x³/3 +3x + 2

y = 1/f(x) = 1/([1-cos(πx)]/π + x³/3 +3x + 2)

The tangent to y is thus dy/dx

dy/dx = d1/([1-cos(πx)]/π + x³/3 +3x + 2)/dx

dy/dx = -([1-cos(πx)]/π + x³/3 +3x + 2)⁻²(sin(πx) + x² +3)

at x = 0,

dy/dx = -([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)⁻²(sin(π × 0) + 0² +3)

dy/dx = -([1-cos(0)]/π + 0 + 0 + 2)⁻²(sin(0) + 0 +3)

dy/dx = -([1 - 1]/π + 0 + 0 + 2)⁻²(0 + 0 +3)

dy/dx = -(0/π + 2)⁻²(3)

dy/dx = -(0 + 2)⁻²(3)

dy/dx = -(2)⁻²(3)

dy/dx = -3/4

At x = 0,

y = 1/([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)

y = 1/([1-cos(0)]/π + 0 + 0 + 2)

y = 1/([1 - 1]/π + 2)

y = 1/(0/π + 2)

y = 1/(0 + 2)

y = 1/2

So, the equation of the tangent at (0, 1/2) is

\frac{y - \frac{1}{2} }{x - 0} = -\frac{3}{4}  \\y - \frac{1}{2} = -\frac{3}{4}x\\y = -\frac{3}{4}x + \frac{1}{2}

c. If g(x) = f (√(3x² + 4). Find g'(2)

g(x) = f (√(3x² + 4) = [1-cos(π√(3x² + 4)]/π + √(3x² + 4)³/3 +3√(3x² + 4) + 2

g'(x) = [3xsinπ√(3x² + 4) + 18x(3x² + 4) + 9x]/√(3x² + 4)

g'(2) = [3(2)sinπ√(3(2)² + 4) + 18(2)(3(2)² + 4) + 9(2)]/√(3(2)² + 4)

g'(2) = [6sinπ√(12 + 4) + 36(12 + 4) + 18]/√12 + 4)

g'(2) = [6sinπ√(16) + 36(16) + 18]/√16)

g'(2) = [6sin4π + 576 + 18]/4)

g'(2) = [6 × 0 + 576 + 18]/4)

g'(2) = [0 + 576 + 18]/4)

g'(2) = 594/4

g'(2) = 148.5

d. If h be the inverse function of f. Find h' (2)

If h(x) = f⁻¹(x)

then h'(x) = 1/f'(x)

h'(x) = 1/(sin(πx) + x² +3)

h'(2) = 1/(sin(π2) + 2² +3)

h'(2) = 1/(sin(2π) + 4 +3)

h'(2) = 1/(0 + 4 +3)

h'(2) = 1/7

7 0
3 years ago
I need help please.
Norma-Jean [14]

Answer:

No

Step-by-step explanation:

7 0
3 years ago
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